(i) Let f(x) = -x. Accordingly, f(x+h)=-(x+h)
By first principle,
f'(x)=limh→0 hf(x+h)−f(x)
=limh→0 h−(x+h)−(−x)
=limh→0 h−x−h+x
=limh→0 h−h
=limh→0(-1)=-1
(ii) Let f(x)=(-x)-1 = −x1 = −x1 Accordingly, f(x+h)=−x+h1
By first principle,
f'(x)=limh→0 hf(x+h)−f(x)
=limh→0 h1[−x+h1−(x−1)]
=limh→0h1[x+h−1+x1]
=limh→0h1[x(x+h)−x+(x+h)]
=limh→0h1[x(x+h)−x+x+h]
=limh→0h1[xh(x+h)]
=limh→0 x1(x+h)
= x1.x - x21
(iii) Let f(x) = sin (x + 1). Accordingly, f(x+h)=sin(x+h +1)
By first principle,
f'(x)=limh→0 hf(x+h)−f(x)
=limh→0 h1[sin(x+h+1) -sin(x+1)]
=limh→0 1h1[2cos(x+h+1+x+21) sin(x+h+1-x-1)]
=limh→0 h1[ 2cos (2x + h+22) sin(2h)]
=limh→0 h1[ 2cos (22x+h+2) sin(2h)/(2h)]
=cos(22x+0+2).1 [limh→0 xsinx =1]
=cos(x+1)
(iv) letf(x) cos(x-8π). Accordingly, f(x+h)=cos(x+h -8π)
By first principle,
f'(x)=limh→0 hf(x+h)−f(x)
=limh→0h1[cos(x+h-8π)- cos(x-8π)]
=limh→0h1[2sin(2x+h−8π−x+8π)−2sin(x+h−8π+−8π)]
=limh→0h1[2−2sin(2x+h−4π) sin2h]
=limh→0[-sin(22x+h−4π) 2hsin2h]
=-sin(22x+0−4π) .1
=-sin(x-8π)